Nevada, Ohio

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Nevada (pronounced nah-VAY-da) is a village in Wyandot County, Ohio, United States. The population was 814 at the 2000 census.



Nevada is located at 40°49′1″N 83°7′54″W / 40.81694°N 83.13167°W / 40.81694; -83.13167 (40.816867, -83.131664)[3].

According to the United States Census Bureau, the village has a total area of 1.0 square miles (2.7 km²), all of it land.


As of the census[2] of 2000, there were 814 people, 313 households, and 215 families residing in the village. The population density was 788.8 people per square mile (305.1/km²). There were 328 housing units at an average density of 317.8/sq mi (123.0/km²). The racial makeup of the village was 97.42% White, 0.37% African American, 0.12% Native American, 0.25% Asian, 0.12% Pacific Islander, 0.74% from other races, and 0.98% from two or more races. Hispanic or Latino of any race were 1.11% of the population.

There were 313 households out of which 33.5% had children under the age of 18 living with them, 56.5% were married couples living together, 8.6% had a female householder with no husband present, and 31.0% were non-families. 24.9% of all households were made up of individuals and 13.1% had someone living alone who was 65 years of age or older. The average household size was 2.60 and the average family size was 3.13.

In the village the population was spread out with 27.4% under the age of 18, 8.2% from 18 to 24, 30.1% from 25 to 44, 21.4% from 45 to 64, and 12.9% who were 65 years of age or older. The median age was 34 years. For every 100 females there were 98.1 males. For every 100 females age 18 and over, there were 91.9 males.

The median income for a household in the village was $34,706, and the median income for a family was $41,250. Males had a median income of $32,159 versus $21,181 for females. The per capita income for the village was $15,395. About 2.7% of families and 4.9% of the population were below the poverty line, including 3.8% of those under age 18 and 3.0% of those age 65 or over.

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